Question

18 gm of glucose (C$_6$H$_{12}$O$_6$) is dissolved in 1 kg of water in a saucepan. At what temperature will the water boil at 1.013 bar pressure ? K$_b$ for water is 0.52 K kg mol$^{-1}$.

Hint:

Always check the pressure units. 1.013 bar is standard atmospheric pressure, so pure water boils at 100$^\circ$C.

Answer 1

Given:
Mass of solute (Glucose), $w_2 = 18$ g.
Molar mass of Glucose (C$_6$H$_{12}$O$_6$), $M_2 = 180$ g/mol.
Mass of solvent (Water), $w_1 = 1$ kg.
$K_b = 0.52$ K kg mol$^{-1}$.
Step 1: Calculate moles of glucose ($n_2$):
$n_2 = \frac{18}{180} = 0.1$ mol.
Step 2: Calculate Molality ($m$):
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1}{1} = 0.1$ m.
Step 3: Calculate Elevation in boiling point ($\Delta T_b$):
$\Delta T_b = K_b \times m = 0.52 \times 0.1 = 0.052$ K.
Step 4: Calculate Boiling Point of solution ($T_b$):
Pure water boils at 373.15 K (100$^\circ$C) at 1.013 bar.
$T_b = T_b^0 + \Delta T_b = 373.15 + 0.052 = 373.202$ K.
(In Celsius: $100 + 0.052 = 100.052^\circ$C).